Physics, asked by nanafiker22, 4 months ago

the angular velocity of a rotating body increase uniformly from 100rad/sec to 200rad/sec in 20sec,how many revolutions will the body execute during this time

Answers

Answered by Ekaro
17

Given :

The angular velocity of a rotating body increase uniformly from 100rad/s to 200rad/s in 20s.

To Find :

No. of revolutions completed by the body.

Solution :

❖ First of all we need to find angular acceleration of the body.

  • It is defined as the rate of change of angular velocity of body.

Formula : α = (ω' - ω) / t

» α denotes angular acceleration

» ω' denotes final angular velocity

» ω denotes initial angular velocity

» t denotes time

By substituting the given values;

➙ α = (200 - 100) / 20

➙ α = 100/20

➙ α = 5 rad/s²

Now let's calculate angular distance covered by the body.

ω'² - ω² = 2αθ

➠ (200)² - (100)² = 2(5)θ

➠ 40000 - 10000 = 10 θ

➠ 30000 = 10 θ

➠ θ = 3000 rad

No. of revolutions :

➝ n = θ / 2π

➝ n = 3000/2(3.14)

n ≈ 477 revolutions

Answered by Anonymous
2

Given :

The angular velocity of a rotating body increase uniformly from 100rad/s to 200rad/s in 20s.

To Find :

No. of revolutions completed by the body.

Solution :

❖ First of all we need to find angular acceleration of the body.

It is defined as the rate of change of angular velocity of body.

Formula : α = (ω' - ω) / t

» α denotes angular acceleration

» ω' denotes final angular velocity

» ω denotes initial angular velocity

» t denotes time

By substituting the given values;

➙ α = (200 - 100) / 20

➙ α = 100/20

➙ α = 5 rad/s²

Now let's calculate angular distance covered by the body.

➠ ω'² - ω² = 2αθ

➠ (200)² - (100)² = 2(5)θ

➠ 40000 - 10000 = 10 θ

➠ 30000 = 10 θ

➠ θ = 3000 rad

No. of revolutions :

➝ n = θ / 2π

➝ n = 3000/2(3.14)

➝ n ≈ 477 revolutions

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